CHEMISTRY COMMUNITY

2c_britneyly

I haven't seen a practice final on here, but I think Lyndon will post his soon. There is another review for thermochem/dynamics. You can find it if you search "Nathan's Review Session Problems + Solutions"

2c_britneyly

With elementary steps there's no ambiguity. If the equation is A+A--> B, then you know it requires two moles of A for B to form. If they only give you the overall reaction with no elementary steps, then there could be an intermediate you didn't know about.

2c_britneyly

If you're told the cell does work, cell potential should always be positive. You flip the reduction half reactions and go with whatever gives you the higher voltage. However, if you're not told the cell does work and given a cell diagram, the anode (the species being oxidized) is on the left, and th...

2c_britneyly

Lavelle's outline on his website for chemical kinetics only has the equations listed for zeroth, first, and second order calculations, so I'm assuming this is all we'll need to know for the final. He introduced the idea of higher order reactions, but it is beyond the scope of this class.

2c_britneyly

This question is basically asking you to compare their rates, which can be determined by looking at their stoichiometric coefficients. In this question, you're given the equation A+3B --> 2C. You know that A will be decreasing 3 times slower than B because only one mole of A is decomposing for every...

2c_britneyly

When it's not clear if the concentration is doubled, then you need to divide one rate law equation by the other, with the respective experimental numbers for each variable. This only works if everything else stays constant besides the one reactant you are working with. You'll end up with a number on...

2c_britneyly

Yes, it shows the relationship between free energy and cell potential.
This equation can be used to show that when E is positive, deltaG will be negative, thus making the reaction spontaneous.

2c_britneyly

E is the cell potential at any stage of a reaction, so the reaction is not necessarily at equilibrium. E naught is cell potential under standard conditions.

2c_britneyly

The Nernst Equation shows us how cell potential is dependent on concentration. We use this equation when we want to calculate the cell potential using the concentration of our reactants and products. Vice versa, we would also be able to calculate the concentration of reactant or product when given t...

2c_britneyly

It's just convention so people can easily identify the anode and cathode. There's no scientific reason behind it.

2c_britneyly

Yes, you use the enthalpy and entropy given in the tables. Check your units. They usually give enthalpy in kJ and entropy in J, so it's possible you forgot to convert.

2c_britneyly

It will be spontaneous at a low temperature.

2c_britneyly

The sign of enthalpy is determined with respect to the system. Therefore, when ΔH is negative, you know heat is decreasing with respect to the system. This is the conventional way of writing signs unless otherwise stated.

2c_britneyly

I highly recommend attending UA sessions since they create worksheets a bit more on the challenging side. Although the worksheets aren't always accurate reflections of the test, you come out with a deeper understanding of the material.

2c_britneyly

It helps to think about it using the ideal gas law, PV=nRT. If moles and temperature remain the same, then you know the entire right hand of the equation is a constant. Say, for example, you increase volume. Then you know since nothing else is changing, pressure has to go down in order for the right...

2c_britneyly

ΔS=q/T, so the higher the temperature, the smaller the change in entropy for the same amount of heat released or absorbed. ΔS=nRlnV 2 /V 1 , so if there is a decrease in volume the ln term will become negative, and there will be a decrease in entropy. If volume increases, the ln term becomes positiv...

2c_britneyly

When ΔV=0, ΔU=q+w=q-PΔV. Therefore when we substitute ΔV=0, the term -PΔV turns into zero, and we're left with ΔU=q.
When ΔP=0, then ΔU=q_p+w. Heat at a constant pressure is enthalpy, thus the equation will be ΔU=ΔH-PΔV

2c_britneyly

The equation ΔS=q/T is applicable to an isothermal, reversible reaction. The other equation ΔS=k B lnW is used to calculate residual entropy, meaning the system is cooled to nearly 0K so molecules are barely moving. We are calculating only the entropy of molecules, not a reaction with heat. A good w...

2c_britneyly

I don't think it has anything to do with entropy, this is just calculating work with different conditions. The basis of deriving a new equation is due to the slowly decreasing external pressure when the system is expanding. Because P is not constant, we derive a new P by rearranging PV=nRT to P=nRT/...

2c_britneyly

Extensive properties are dependent on the amount of substance, but intensive properties are not. Specific heat capacities are intensive because they're "specific for 1 gram, " while extensive properties would be heat capacities (but not specific to 1 gram of the substance).

2c_britneyly

I'm pretty sure they will give us a table of bond enthalpies if we're required to do those calculations. I would say the only ones you need to know are the basic ones such as C-C bonds, which you'll probably memorize while doing the homework.

2c_britneyly

Heat capacity is the amount of energy it takes to raise the temperature of a substance by 1 degree Celsius. Molar heat capacity is the amount of energy it takes to raise the temperature of a substance by 1 degree Celsius per mole of a substance. These two are essentially the same thing, but molar he...

2c_britneyly

Any time there is a change in temperature then K_w is subject to change. For the purpose of this class we assume K_w=10^-14

2c_britneyly

When filling out the ICE table, your initial amt. of benzoic acid is .110 M, and the equilibrium concentration of the products (more importantly, equilibrium concentration of H+) is (.024 x .110 M). Since you know [H+], you can find pH. Then set up an equilibrium expression to find K a . It should b...

2c_britneyly

Yes, if you're dealing with a reaction that involves only gases it is more correct to use P_gas, but if you're calculating for a reaction that involves an aqueous solution, you should use hard brackets to indicate K_c.

2c_britneyly

Both flask 3 and 4 have the same mole ratios, but I'm assuming they're asking for when the system first reaches equilibrium, which would be flask 3.

2c_britneyly

Because they tell you it reaches equilibrium after 54.5% of the molecules dissociate, you can set up an ICE table, with the initial as .1 and the change as -(.545x.1). That will give you the final partial pressure of X2 and you can find the final pressure of 2X by doubling the change of X2 due to th...

2c_britneyly

Your first term of the equation should be 4X² rather than 2X². This is just a mathematical mistake. It should be (2X)² rather than 2(X)².

2c_britneyly

Products are not always more stable than reactants. In his example the K value just happened to be large, meaning the reaction favors the products. Reactants can also be more stable than reactants, which would mean a smaller K value.

2c_britneyly

For this example, the mole ratios of ADP to AD and ADP to P_i are 1, so the change in moles of products is the same as change in moles of reactants. If the mole ratios were different, you would have to multiply by stoichiometric coefficients before finding equilibrium concentrations.

2c_britneyly

Yes, you have to get rid of spectator ions by writing a net ionic equation first. I'm assuming for this course we only need to identify simple ionic compounds, such as AgCl.

2c_britneyly

Compounds that have hydrogen atoms written separately in the formula usually indicate they are the terminal atom and most likely dissociate in water. A good example is CH3COOH. They could have written the compound as CH4COO, but writing it the former way indicates the H is attached at another locati...

2c_britneyly

Yes, we still have homework due this week in discussion.

2c_britneyly

If they give you a table of the dissociation constants, then you can easily determine which ones are stronger or weaker by looking at the numbers.
Generally, you just have to memorize the strong acids, such a H2SO4, HCl, etc.

2c_britneyly

Maybe there's a typo in the answer key/version you have? I have the 7th ed. and it says the answer is bent.

2c_britneyly

Because it's not listed on the Ch4 Outline, I would say we don't need to calculate exact bond lengths for the test. I would check with a TA or Lavelle tomorrow to be sure.

2c_britneyly

Li+ and Mg2+ have higher polarizabilities, respectively. In (a), both Li+ and Be2+ have the same number of electrons. However, because Be 2+ has more protons, there is a higher nuclear charge, and electrons are drawn in more closely, making the cation smaller and not as likely to lose electrons. Li+...

2c_britneyly

Test 3 begins this week in your discussion. On the website it says the test will cover 2D to 2F, so that would be VSEPR, hybridization, shapes/bond angles, and properties of bonds (eg bond length, polarizability which you can find in 2D).

2c_britneyly

The arrows should point in the direction of the atom with higher electronegativity, indicating an area with higher electron density

2c_britneyly

The general rule is that the least electronegative atom is placed in the center. So if you're drawing a Lewis structure that has carbon and oxygen atoms, carbon will be the central atom. However, for elements in row 3 and beyond with available space in the d shell, they can have expanded octets and ...

2c_britneyly

Yes, I would do that to be safe on a test. If they're resonance structures that have the same number of each type of bond then you could probably say all of them have the same number of pi and sigma bonds.

2c_britneyly

You generally only consider charge as a factor when the ions are positively charged because that's when atoms are seeking more electrons.

2c_britneyly

It helps if you start by memorizing the basic models where the central atom is bonded to all atoms, such as octahedral. From there, you can replace each bonding atom with lone pairs of electrons and determine the shape. So for an octahedral, you can replace one bonding atom with a lone pair, and tha...

2c_britneyly

Yes, you can also think of it in terms of atomic radius. Since phosphorus has more electrons, its electron cloud (and as a result its atomic radius) is larger than nitrogen. Since distance between the two atoms is greater, we know the bond length will also be greater.

2c_britneyly

Dipole-dipole interactions occur between two molecules that already have dipole moments, such as an interaction between 2 HF molecules. In the case of dipole induced dipole, a molecule with a dipole moment will induce a dipole moment in another atom without a dipole moment by distorting its electron...

2c_britneyly

Yes, the more electrons two atoms share, the closer they will be drawn to one another, thus decreasing bond length.

2c_britneyly

Larger atoms have more electrons in higher energy levels. As a result, electrons are not bound as tightly to the nucleus and each atom will have a larger electron cloud. Because electrons are loosely attracted to the nucleus, it's easier for them to shift around, so electron clouds are more easily d...

2c_britneyly

The most likely charge is 3+ because atoms always try to reach their most stable state. For gallium, it's easier for it to become stable by giving up three electrons rather than receiving 5 electrons to complete its 4p shell. That's why most of the elements on the left side of the periodic table giv...

2c_britneyly

You use De Broglie's equation for everything with mass (electron, proton, everyday objects). You use E=hv for a photon because it has no resting mass.

2c_britneyly

You use De Broglie's equation for objects with mass. This includes everyday objects, electrons, and protons. You use E=hv when you are doing calculations for a photon since a photon has no resting mass.

2c_britneyly

You would be solving for the uncertainty in velocity. h is a constant. 350 pm would be delta x, or the uncertainty in position.

2c_britneyly

Schrodinger's wave equation gives us electron probability densities, and when we graph them it gives us orbital shapes. However, some orbitals have areas where the electron probability density is zero, meaning you will never find an electron there. This results in a nodal plane that consists of mult...

2c_britneyly

Does anyone remember what Lavelle said the smallest detectable wavelength is?

2c_britneyly

To do this we need to know quantum numbers, which is another way to describe orbitals. The variable "n" represents principal energy levels and can take the values of 1,2,3,4,etc. "l" represents subshells (s,p,d,f) and takes the values of 0,1,2,...,n-1. "m" represents or...

2c_britneyly

Because of the math behind Schrodinger's wave function equation, there are points where the calculated electron probability density is zero, indicating the presence of a nodal plane. The most important thing to know about them is you will never find an electron there. In other orbitals such as dz2 a...

2c_britneyly

From part (a), you know that the agents should use the low-intensity GaN laser. For part (b), you want to use the equation E photon -work function=E k to find the kinetic energy emitted. Since you know the wavelength, you can use the equation E=hv to find energy per photon. You are given the work fu...

2c_britneyly

If you look at example 1A.2 in the book, your work process should be exactly like the example but backwards. Because you are given wavelength, you can solve for frequency and plug it back into the Rydberg equation to solve for n₂

2c_britneyly

I think Dr. Lavelle explained this using an example of hurdles. If someone doesn't have enough energy, then they will not be able to successfully jump over the hurdle. Similarly, if a photon does not have enough energy, then we say it does not meet the threshold energy, and therefore cannot remove a...

2c_britneyly

Same, I solved it and ended up with 150 picometers too.

2c_britneyly

Hi Nadeen! You can solve these by using the equation M(initial)*V(initial)=M(final)*V(final). This problem is essentially asking you to find the initial volume used to make these dilutions. First, you want to determine what values you are already given. You are given the final number of moles, which...

2c_britneyly

Your method sounds correct, and the other people who previously answered got 76.5% too. The most likely reason for the difference in percentage is that the book may have used a different molar mass for the compound N2H4.

2c_britneyly

Zeros between digits other than zero are always considered significant. The way I like to think of it is that those zeros can't be removed without changing the value of the number.

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